Temperature and Temperature scales Questions in English

(B) The relationship between the Fahrenheit scale $(F)$ and the Kelvin scale $(K)$ is given by the formula: $\frac{F - 32}{9} = \frac{K - 273}{5}$.
Given that the temperature is $X$ on both scales,we substitute $F = X$ and $K = X$ into the equation:
$\frac{X - 32}{9} = \frac{X - 273}{5}$.
Cross-multiplying gives: $5(X - 32) = 9(X - 273)$.
Expanding the terms: $5X - 160 = 9X - 2457$.
Rearranging the terms to solve for $X$: $9X - 5X = 2457 - 160$.
$4X = 2297$.
$X = \frac{2297}{4} = 574.25$.
Therefore,the value of $X$ is $574.25$.

(C) An ideal thermometer must reach thermal equilibrium with the body whose temperature is being measured.
To ensure the measurement is accurate,the thermometer should absorb or release as little heat as possible from the body.
If the thermometer has a large heat capacity,it will draw a significant amount of heat from the body,thereby changing the body's original temperature.
Therefore,to minimize the disturbance to the system,an ideal thermometer should have a small heat capacity.

(C) The relationship between a faulty scale $(X)$ and the Celsius scale $(C)$ is given by the formula: $\frac{X - L_X}{U_X - L_X} = \frac{C - L_C}{U_C - L_C}$.
Here,the lower fixed point of the faulty scale $L_X = -10^\circ$ and the upper fixed point $U_X = 110^\circ$.
For the Celsius scale,the lower fixed point $L_C = 0^\circ$ and the upper fixed point $U_C = 100^\circ$.
Given the temperature on the faulty scale $X = 62^\circ$,we substitute the values into the formula:
$\frac{62 - (-10)}{110 - (-10)} = \frac{C - 0}{100 - 0}$
$\frac{62 + 10}{110 + 10} = \frac{C}{100}$
$\frac{72}{120} = \frac{C}{100}$
$C = \frac{72}{120} \times 100 = 0.6 \times 100 = 60^\circ C$.

(C) By definition,one calorie is the amount of heat energy required to raise the temperature of $1\ g$ of pure water from $14.5\ ^oC$ to $15.5\ ^oC$ at a standard atmospheric pressure of $760\ mm$ of $Hg$.

(C) The change in temperature on the Celsius scale is given by $\Delta T_C = 90\ ^oC - 30\ ^oC = 60\ ^oC$.
For the Fahrenheit scale,the relation between the change in temperature is $\Delta T_F = \frac{9}{5} \Delta T_C$.
Substituting the value: $\Delta T_F = \frac{9}{5} \times 60\ ^oC = 108\ ^oF$.
For the Kelvin scale,the change in temperature is equal to the change in temperature on the Celsius scale,because the size of one degree Celsius is equal to the size of one Kelvin unit.
Therefore,$\Delta T_K = \Delta T_C = 60\ K$.
Thus,the change is $108\ ^oF$ and $60\ K$.

(C) Let $T_{C}$ be the correct temperature on the Celsius scale.
The formula relating a faulty scale to the Celsius scale is given by:
$\frac{T_{C} - 0}{100 - 0} = \frac{X - X_{L}}{X_{U} - X_{L}}$
Where:
$X = 59^{\circ}$ (reading on the faulty thermometer)
$X_{L} = 5^{\circ}$ (lower fixed point)
$X_{U} = 95^{\circ}$ (upper fixed point)
Substituting the values:
$\frac{T_{C}}{100} = \frac{59 - 5}{95 - 5}$
$\frac{T_{C}}{100} = \frac{54}{90}$
$\frac{T_{C}}{100} = \frac{6}{10} = 0.6$
$T_{C} = 0.6 \times 100 = 60^{\circ}$

(B) The formula to convert temperature between two scales is given by: $\frac{X - X_{LFP}}{X_{UFP} - X_{LFP}} = \frac{C - C_{LFP}}{C_{UFP} - C_{LFP}}$.
Here,the faulty thermometer has a Lower Fixed Point $(X_{LFP})$ of $5^{\circ}$ and an Upper Fixed Point $(X_{UFP})$ of $95^{\circ}$.
The Celsius scale has a Lower Fixed Point $(C_{LFP})$ of $0^{\circ}$ and an Upper Fixed Point $(C_{UFP})$ of $100^{\circ}$.
Given the reading on the faulty thermometer $X = 59^{\circ}$.
Substituting the values: $\frac{59 - 5}{95 - 5} = \frac{C - 0}{100 - 0}$.
$\frac{54}{90} = \frac{C}{100}$.
$C = \frac{54}{90} \times 100 = 0.6 \times 100 = 60^{\circ}$.

(C) For any temperature scale,the ratio of the difference between the reading and the melting point (ice point) to the difference between the boiling point (steam point) and the melting point is constant.
$\frac{\text{Reading}_X - MP_X}{BP_X - MP_X} = \frac{\text{Reading}_Y - MP_Y}{BP_Y - MP_Y}$
Given for thermometer $X$: $MP_X = 15^{\circ}$,$BP_X = 75^{\circ}$,Reading $= 60^{\circ}$.
Given for thermometer $Y$: $MP_Y = 25^{\circ}$,$BP_Y = 125^{\circ}$.
Substituting the values:
$\frac{60 - 15}{75 - 15} = \frac{y - 25}{125 - 25}$
$\frac{45}{60} = \frac{y - 25}{100}$
$\frac{3}{4} = \frac{y - 25}{100}$
$y - 25 = 75$
$y = 100^{\circ}$

(B) The initial temperature of boiling water is $100\,^{\circ}C$ or $212\,^{\circ}F$.
The final temperature in Fahrenheit is $F = 140\,^{\circ}F$.
Using the conversion formula $\frac{C}{100} = \frac{F-32}{180}$,we find the final temperature in Celsius:
$\frac{C}{100} = \frac{140-32}{180} = \frac{108}{180} = 0.6$
$C = 60\,^{\circ}C$.
The fall in temperature on the Centigrade scale is $\Delta C = 100\,^{\circ}C - 60\,^{\circ}C = 40\,^{\circ}C$.

(C) The heat required to raise the temperature of a substance is given by $H = ms\Delta\theta$.
For $1 \text{ g}$ of water,$m$ and $s$ are constant.
$H_C = ms(1^{\circ}C)$.
$H_K = ms(1 \text{ K})$. Since a change of $1 \text{ K}$ is equal to a change of $1^{\circ}C$,$H_K = ms(1^{\circ}C) = H_C$.
$H_F = ms(1^{\circ}F)$. Since a change of $1^{\circ}F$ is equal to $(5/9)^{\circ}C$,$H_F = ms(5/9)^{\circ}C$.
Comparing the values,$H_C = H_K$ and since $1 > 5/9$,$H_C > H_F$.
Therefore,$H_K = H_C > H_F$.

(C) The relationship between any two temperature scales is given by the formula: $\frac{T - LFP}{UFP - LFP} = \text{constant}$.
For the Kelvin scale $(K)$: $LFP = 273\,K$,$UFP = 373\,K$.
For the $Y$ scale: $LFP = -160^{\circ}Y$,$UFP = -50^{\circ}Y$.
Setting the ratios equal:
$\frac{340 - 273}{373 - 273} = \frac{Y - (-160)}{-50 - (-160)}$
$\frac{67}{100} = \frac{Y + 160}{110}$
$Y + 160 = \frac{67 \times 110}{100}$
$Y + 160 = 73.7$
$Y = 73.7 - 160 = -86.3^{\circ}Y$.

(A) For a linear scale,the temperature $T$ can be related to the reading $x$ by the formula:
$T = \frac{x - x_{ice}}{x_{steam} - x_{ice}} \times 100^\circ C$
Given:
$x_{steam} = x_0$
$x_{ice} = x_0/3$
$x = x_0/2$
Substituting these values into the formula:
$T = \frac{x_0/2 - x_0/3}{x_0 - x_0/3} \times 100^\circ C$
$T = \frac{(3x_0 - 2x_0)/6}{(3x_0 - x_0)/3} \times 100^\circ C$
$T = \frac{x_0/6}{2x_0/3} \times 100^\circ C$
$T = \frac{x_0}{6} \times \frac{3}{2x_0} \times 100^\circ C$
$T = \frac{1}{4} \times 100^\circ C = 25^\circ C$

(D) The density of water is maximum at $4^{\circ}C$.
To convert this temperature to Fahrenheit,we use the formula:
$\frac{T_{C}-0}{100-0} = \frac{T_{F}-32}{212-32}$
Substituting $T_{C} = 4^{\circ}C$:
$\frac{4}{100} = \frac{T_{F}-32}{180}$
$0.04 \times 180 = T_{F} - 32$
$7.2 = T_{F} - 32$
$T_{F} = 32 + 7.2 = 39.2^{\circ}F$
Thus,the maximum density of $H_2O$ occurs at $39.2^{\circ}F$.

(C) The relationship between the Fahrenheit scale $(F)$ and the Kelvin scale $(K)$ is given by the formula: $\frac{F-32}{9} = \frac{K-273}{5}$.
Given that the temperature is $x$ on both scales,we substitute $F = x$ and $K = x$ into the equation:
$\frac{x-32}{9} = \frac{x-273}{5}$.
Cross-multiplying the terms,we get: $5(x-32) = 9(x-273)$.
Expanding the brackets: $5x - 160 = 9x - 2457$.
Rearranging the terms to solve for $x$: $2457 - 160 = 9x - 5x$.
$2297 = 4x$.
$x = \frac{2297}{4} = 574.25$.
Therefore,the value of $x$ is $574.25$.

(A) The relationship between Fahrenheit $(T_F)$ and Celsius $(T_C)$ scales is given by the formula: $\frac{T_F - 32}{180} = \frac{T_C}{100}$.
Let the reading on the centigrade thermometer be $\theta$. According to the problem,the reading on the Fahrenheit thermometer is $2\theta$.
Substituting these values into the formula: $\frac{2\theta - 32}{180} = \frac{\theta}{100}$.
Simplifying the equation: $\frac{2\theta - 32}{18} = \frac{\theta}{10}$.
$10(2\theta - 32) = 18\theta$.
$20\theta - 320 = 18\theta$.
$2\theta = 320$.
$\theta = 160$.
Thus,the temperature is $160^{\circ}$.

(D) The density of water is maximum at $4^{\circ}C$.
To convert this temperature to Fahrenheit,we use the formula:
$\frac{C}{5} = \frac{F - 32}{9}$
Substituting $C = 4$:
$\frac{4}{5} = \frac{F - 32}{9}$
$0.8 = \frac{F - 32}{9}$
$7.2 = F - 32$
$F = 39.2^{\circ}F$
Thus,the maximum density of $H_2O$ occurs at $39.2^{\circ}F$.

(B) The relation between the Celsius $(C)$ and Fahrenheit $(F)$ scales of temperature is given by the formula:
$\frac{C}{5} = \frac{F - 32}{9}$
Rearranging this equation to express $C$ in terms of $F$:
$C = \frac{5}{9}F - \frac{160}{9}$
Comparing this with the standard equation of a straight line,$y = mx + c$,where $y$ is the temperature in Celsius,$x$ is the temperature in Fahrenheit,and $m$ is the slope:
Here,$y = C$ and $x = F$.
Thus,the slope $m = 5/9$.
Therefore,the slope of the line $AB$ is $5/9$.

(C) The relationship between two temperature scales is given by the formula: $\frac{X - X_{L}}{X_{U} - X_{L}} = \frac{Y - Y_{L}}{Y_{U} - Y_{L}}$
Here,$X_{L} = 15$,$X_{U} = 75$,$Y_{L} = 25$,and $Y_{U} = 125$.
Given that thermometer $X$ reads $60^{\circ}$,we substitute the values:
$\frac{60 - 15}{75 - 15} = \frac{Y - 25}{125 - 25}$
$\frac{45}{60} = \frac{Y - 25}{100}$
$\frac{3}{4} = \frac{Y - 25}{100}$
$Y - 25 = \frac{3}{4} \times 100 = 75$
$Y = 75 + 25 = 100^{\circ}$

(C) The water starts boiling a second time because when the pressure cooker cools down,the pressure inside decreases.
According to the relationship between pressure and boiling point,a decrease in pressure leads to a decrease in the boiling point $(B.P.)$ of water.
When the lid is removed,the pressure drops to atmospheric pressure,which is lower than the pressure inside the closed cooker.
Since the water is already at a high temperature,this reduction in pressure causes the water to boil again.
The Reason provided is incorrect because impurities generally increase the boiling point of water (elevation of boiling point),not decrease it.

(D) The Assertion is incorrect because the size of a degree in a temperature scale is determined by the number of divisions between the freezing and boiling points of water.
In the Celsius scale $(^oC)$,there are $100$ divisions.
In the Fahrenheit scale $(^oF)$,there are $180$ divisions.
In the Reaumur scale $(^oR)$,there are $80$ divisions.
In the Rankine scale $(^oRa)$,there are $180$ divisions.
Since the Rankine scale has the largest number of divisions for the same temperature interval,its unit size is actually smaller than that of the Fahrenheit scale.
Therefore,the statement that Fahrenheit is the smallest unit is false.
The Reason is correct because Daniel Gabriel Fahrenheit developed the first standardized mercury-in-glass thermometer and the associated Fahrenheit scale in $1724$.

The relation between Kelvin and Celsius scales is given by:
$T_{C} = T_{K} - 273.15$
The relation between Celsius and Fahrenheit scales is given by:
$T_{F} = \frac{9}{5} T_{C} + 32$
For neon $(T_{K} = 24.57\; K)$:
$T_{C} = 24.57 - 273.15 = -248.58^{\circ}C$
$T_{F} = \frac{9}{5}(-248.58) + 32 = -447.44 + 32 = -415.44^{\circ}F$
For carbon dioxide $(T_{K} = 216.55\; K)$:
$T_{C} = 216.55 - 273.15 = -56.60^{\circ}C$
$T_{F} = \frac{9}{5}(-56.60) + 32 = -101.88 + 32 = -69.88^{\circ}F$

(D) The triple point of water on the Kelvin scale is $T_{K} = 273.15\; K$.
Given that the triple point of water on scale $A$ is $200\; A$,we have $200\; A = 273.15\; K$. Thus,$1\; A = \frac{273.15}{200}\; K$.
Given that the triple point of water on scale $B$ is $350\; B$,we have $350\; B = 273.15\; K$. Thus,$1\; B = \frac{273.15}{350}\; K$.
To find the relation between any temperature $T_{A}$ and $T_{B}$ representing the same physical state,we equate them to the Kelvin scale:
$T_{A} \times \left( \frac{273.15}{200} \right) = T_{B} \times \left( \frac{273.15}{350} \right)$.
Dividing both sides by $273.15$,we get:
$\frac{T_{A}}{200} = \frac{T_{B}}{350}$.
Simplifying the ratio:
$T_{A} = \frac{200}{350} T_{B} = \frac{4}{7} T_{B}$.
Therefore,the relation is $T_{A} = \frac{4}{7} T_{B}$ or $T_{A} : T_{B} = 4 : 7$.

(C) The given relation is $R = R_{0} [1 + \alpha (T - T_{0})]$.
At the triple point of water,$T_{0} = 273.16 \; K$ and $R_{0} = 101.6 \; \Omega$.
At the melting point of lead,$T = 600.5 \; K$ and $R = 165.5 \; \Omega$.
Substituting these values into the equation:
$165.5 = 101.6 [1 + \alpha (600.5 - 273.16)]$
$1.629 = 1 + \alpha (327.34)$
$\alpha = \frac{0.629}{327.34} \approx 1.9215 \times 10^{-3} \; K^{-1}$.
Now,for $R = 123.4 \; \Omega$,we find $T$:
$123.4 = 101.6 [1 + 1.9215 \times 10^{-3} (T - 273.16)]$
$1.21457 = 1 + 1.9215 \times 10^{-3} (T - 273.16)$
$0.21457 = 1.9215 \times 10^{-3} (T - 273.16)$
$T - 273.16 = \frac{0.21457}{1.9215 \times 10^{-3}} \approx 111.67$
$T = 111.67 + 273.16 = 384.83 \; K$.
Rounding to the nearest provided option,$T \approx 384.61 \; K$.

(N/A) The triple-point of water is a unique state that occurs at a specific temperature and pressure,making it a highly reproducible standard. The melting point of ice and the boiling point of water are dependent on pressure; therefore,they are not unique unless the pressure is strictly defined,making them less reliable as universal standards.
$(b)$ The other fixed point on the Kelvin absolute scale is absolute zero,which is $0\; K$.
$(c)$ The value $273.16\; K$ is the triple-point of water. The Celsius scale defines $0^{\circ}C$ as the melting point of ice at $1\; atm$ pressure. The difference between the triple-point of water and the melting point of ice is $0.01\; K$. Thus,$0^{\circ}C = 273.16\; K - 0.01\; K = 273.15\; K$. Hence,$t_c = T - 273.15$.
$(d)$ The size of $1\; K$ is equal to $1.8$ times the size of $1^{\circ}F$ (since $100\; K$ interval corresponds to $180^{\circ}F$ interval). The triple-point of water is $273.16\; K$. On a scale where the unit interval is equal to the Fahrenheit scale,the triple-point temperature is $273.16 \times 1.8 = 491.688\; R$ (Rankine scale).

(N/A) From everyday experience,we observe that a hot body has a higher temperature,while a cold body has a lower temperature. For instance,boiling water has a higher temperature than ice.
Temperature is a measure of the 'degree of hotness' or 'coldness' of a body. Heat,on the other hand,is a form of energy that depends on the mass and the nature of the substance. For example,the total heat energy contained in a bucket of water at room temperature is greater than the heat energy in a glass of boiling water,even though the boiling water has a higher temperature.

(N/A) The statement is incorrect.
Heat is a form of energy in transit due to a temperature difference between two bodies.
It is not a property of a body,meaning a body does not 'contain' heat.
Instead,a body contains 'internal energy'.
Temperature is a measure of the average kinetic energy of the molecules in a body.
Therefore,a hot body has a higher temperature than a cold body,but it does not possess a greater 'amount of heat'.
Heat is only defined when energy is transferred from a body at a higher temperature to a body at a lower temperature.

(N/A) Temperature is a relative measure,or indication of hotness or coldness.
$A$ hot utensil is said to have a high temperature and an ice cube is said to have a low temperature.
An object that has a higher temperature than another object is said to be hotter. Hot and cold are relative terms,like tall and short.
We can perceive temperature by touch. However,this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes.
$A$ glass of ice-cold water left on a table on a hot summer day eventually warms up,whereas a cup of hot tea on the same table cools down.
This means that when the temperature of a body (ice-cold water or hot tea in this case) and its surrounding medium are different,heat transfer takes place between the system and the surrounding medium until the body and the surrounding medium are at the same temperature.
In the case of a glass tumbler of ice-cold water,heat flows from the environment to the glass tumbler,whereas in the case of hot tea,it flows from the cup of hot tea to the environment.
Therefore,heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of a temperature difference.
The $SI$ unit of heat energy transferred is expressed in joule $(J)$,while the $SI$ unit of temperature is kelvin $(K)$,and ${ }^{\circ}C$ is a commonly used unit of temperature.

(N/A) Temperature is a physical quantity that measures the degree of hotness or coldness of a body or an environment.
It represents the average kinetic energy of the particles in a substance.
The $SI$ unit of temperature is the Kelvin,denoted by the symbol $K$.

(N/A) measure of temperature is obtained using a thermometer.
Thermometers are calibrated so that a numerical value may be assigned to a given temperature. The commonly used property is the variation of the volume of a liquid with temperature (thermal expansion). Mercury and alcohol are the liquids used in most liquid-in-glass thermometers.
For the definition of any standard scale,two fixed reference points are needed.
The ice point and the steam point of water are two convenient fixed points and are known as the freezing and boiling points.
These two points are the temperatures at which pure water freezes and boils under standard pressure.
The two familiar temperature scales are the Fahrenheit temperature scale and the Celsius temperature scale.
The ice and steam point have values $32^{\circ}F$ and $212^{\circ}F$ respectively on the Fahrenheit scale,and $0^{\circ}C$ and $100^{\circ}C$ on the Celsius scale.
On the Fahrenheit scale,there are $180$ equal intervals between the two reference points,and on the Celsius scale,there are $100$ equal intervals.

(N/A) The relationship between the Celsius and Fahrenheit temperature scales can be derived from a linear graph of Fahrenheit temperature $(t_{F})$ versus Celsius temperature $(t_{C})$.
As shown in the graph,the freezing point of water is $0^{\circ}C$ or $32^{\circ}F$,and the boiling point of water is $100^{\circ}C$ or $212^{\circ}F$.
The equation for this straight line is given by:
$\frac{t_{F} - 32}{212 - 32} = \frac{t_{C} - 0}{100 - 0}$
Simplifying the denominators:
$\frac{t_{F} - 32}{180} = \frac{t_{C}}{100}$
Further simplifying the equation:
$t_{F} - 32 = \frac{180}{100} t_{C}$
$t_{F} - 32 = \frac{9}{5} t_{C}$
$t_{F} = \frac{9}{5} t_{C} + 32$

(N/A) The instrument used to measure temperature is called a $Thermometer$.
For water:
$1$. Freezing point:
- In $^oC$: $0 \ ^oC$
- In $^oF$: $32 \ ^oF$
- In $K$: $273.15 \ K$
$2$. Boiling point:
- In $^oC$: $100 \ ^oC$
- In $^oF$: $212 \ ^oF$
- In $K$: $373.15 \ K$

(N/A) The relationship between the temperature in Fahrenheit $(t_F)$ and the temperature in Celsius $(t_C)$ is derived from the freezing and boiling points of water on both scales.
On the Celsius scale,the freezing point is $0^\circ C$ and the boiling point is $100^\circ C$.
On the Fahrenheit scale,the freezing point is $32^\circ F$ and the boiling point is $212^\circ F$.
The conversion formula is given by:
$t_F = \frac{9}{5} t_C + 32$.

(B) The freezing point of water is $0 ^oC$ or $32 ^oF$. The difference is $32 - 0 = 32$.
The boiling point of water is $100 ^oC$ or $212 ^oF$. The difference is $212 - 100 = 112$.
However,the question asks for the difference between the values on the two scales for these points.
For freezing point: $32 ^oF - 0 ^oC = 32$.
For boiling point: $212 ^oF - 100 ^oC = 112$.
If the question implies the range (interval) difference: The range on the Celsius scale is $100 - 0 = 100$. The range on the Fahrenheit scale is $212 - 32 = 180$.

(N/A) thermometer is a device used to measure the temperature of a system or a body. It works on the principle that certain physical properties of substances change with temperature.
$A$ gas thermometer is a type of thermometer that uses the change in pressure or volume of a gas at a constant volume or constant pressure,respectively,to measure temperature. It is based on the ideal gas law,$PV = nRT$.
Characteristics of a gas thermometer:
$1$. It is highly accurate and is often used as a standard for calibrating other thermometers.
$2$. It has a wide range of temperature measurement,typically from very low temperatures (near absolute zero) to very high temperatures.
$3$. It is based on the properties of an ideal gas,which makes it independent of the specific material properties of the container.
$4$. It is generally bulky and slow to respond to temperature changes compared to liquid-in-glass thermometers.

(N/A) There are three common temperature scales: Celsius $(^{\circ}C)$,Fahrenheit $(^{\circ}F)$,and Kelvin $(K)$.
$(1)$ Relation between Celsius and Kelvin scales:
If $T_{C}$ is the temperature in Celsius and $T$ is the temperature in Kelvin,the relationship is given by:
$T = T_{C} + 273.15$
$(2)$ Relation between Celsius and Fahrenheit scales:
If $T_{F}$ is the temperature in Fahrenheit and $T_{C}$ is the temperature in Celsius,the relationship is given by:
$T_{F} = \frac{9}{5} T_{C} + 32$
$(3)$ General relation:
The general relation between these scales is given by the formula:
$\frac{T_{C} - 0}{100} = \frac{T_{F} - 32}{180} = \frac{T - 273.15}{100}$
These scales are compared based on fixed points like the ice point (freezing point of water) and the steam point (boiling point of water) at atmospheric pressure. The Kelvin scale is an absolute scale where $0 \ K$ is the absolute zero temperature,which is the lowest possible temperature.

(C) The absolute temperature scale is defined by the $Kelvin$ scale.
In this scale,the base or the zero point is $0 \ K$,which corresponds to absolute zero,the theoretical temperature at which all molecular motion ceases.
Therefore,the base of the absolute temperature scale is the $Kelvin$ scale.

(NO) No,a negative value on the Kelvin scale is not possible.
According to the kinetic theory of gases,the temperature of a system is directly proportional to the average kinetic energy of its molecules.
The absolute zero temperature $(0 \ K)$ corresponds to the state where the kinetic energy of the molecules becomes zero,meaning all molecular motion ceases.
Since kinetic energy cannot be negative,the temperature on the Kelvin scale cannot be less than $0 \ K$.

(B) The relationship between the Celsius scale $(C)$ and the Fahrenheit scale $(F)$ is given by the formula: $F = \frac{9}{5}C + 32$.
Absolute zero is defined as $0 \ K$,which is equal to $-273.15 \ °C$.
Substituting $C = -273.15$ into the formula:
$F = \frac{9}{5}(-273.15) + 32$
$F = 1.8 \times (-273.15) + 32$
$F = -491.67 + 32$
$F = -459.67 \ °F$.
Therefore,the absolute zero temperature on the Fahrenheit scale is $-459.67 \ °F$.

(N/A) Temperature is a physical quantity that expresses the degree of hotness or coldness of a body.
It is a measure of the average kinetic energy of the particles in a substance.
When two bodies are in thermal contact,heat flows from the body at a higher temperature to the body at a lower temperature until thermal equilibrium is reached.
The $SI$ unit of temperature is Kelvin $(K)$.

(D) Given:
Resistance at ice point $(R_0) = 5\; \Omega$
Resistance at steam point $(R_{100}) = 5.39\; \Omega$
Resistance in hot bath $(R_t) = 5.795\; \Omega$
The formula for temperature $(t)$ in a platinum resistance thermometer is given by:
$t = \frac{R_t - R_0}{R_{100} - R_0} \times 100$
Substituting the values:
$t = \frac{5.795 - 5}{5.39 - 5} \times 100$
$t = \frac{0.795}{0.39} \times 100$
$t = 2.03846 \times 100$
$t \approx 203.85^{\circ}C$

(A) The density of water is maximum at $4^\circ C$.
To convert Celsius to Fahrenheit, use the formula: $F = C \times (9/5) + 32$.
$F = 4 \times (1.8) + 32 = 7.2 + 32 = 39.2^\circ F$.
To convert Celsius to Kelvin, use the formula: $K = C + 273.15$.
$K = 4 + 273.15 = 277.15 K$.
Thus, the values are $4^\circ C$, $39.2^\circ F$, and $277.15 K$.

(N/A) No,negative temperature is not possible on the Kelvin scale. The Kelvin scale is an absolute temperature scale where the lowest possible temperature is absolute zero $(0 \ K)$. According to the third law of thermodynamics,it is impossible to reach absolute zero,and temperatures below this point cannot exist because the kinetic energy of particles cannot be less than zero.

(N/A) On the Kelvin scale,there is only one fixed point defined by international agreement,which is the triple point of water.
The value of the triple point of water on the Kelvin scale is $273.16 \ K$,which corresponds to $0.01^{\circ}C$ on the Celsius scale.

(A) Absolute zero temperature is the theoretical temperature at which the entropy of a perfect crystal is zero and all molecular motion ceases. It is defined as $0 \ K$ on the Kelvin scale,which is equivalent to $-273.15 \ °C$.

(N/A) The absolute Kelvin scale has only one fixed point,which is the triple point of water.
Its value is defined as $T_{tr} = 273.16 \ K$ or $0.01^{\circ}C$.

(N/A) The triple point is defined as the specific temperature and pressure at which all three states of matter—solid,liquid,and gas—coexist in thermodynamic equilibrium.

(A) For an ideal gas, the graph of pressure $P$ versus temperature $T$ is linear.
By extending this linear graph backwards, the point at which it intersects the temperature axis is called absolute zero temperature.
Its value is $-273.15^{\circ} C$.

(NONE) False. The correct conversion formula is $t^{\circ}F = \frac{9}{5}t^{\circ}C + 32$.
$(b)$ False. Absolute zero temperature is $-273.15^{\circ}C$.
$(c)$ False. The density of water is maximum at $4^{\circ}C$,which means its volume is minimum at this temperature.
$(d)$ False. Heat flows from a higher temperature to a lower temperature. If heat is transferred from the cold end to the hot end,the temperature gradient would be positive in the direction of heat flow.

(A) diathermic wall allows the exchange of heat between systems,whereas an adiabatic wall prevents the exchange of heat. For a thermometer to measure the temperature of a system,it must be in thermal equilibrium with that system. Therefore,the bulb of a thermometer must be made of a diathermic wall to allow heat transfer.